AMC 8 · 2022 · #4
Grade 8 geometry-2dProblem
The letter M in the figure below is first reflected over the line q and then reflected over the line p. What is the resulting image?
Pick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: An upright letter $M$ sits in the first quadrant of a coordinate plane. Line $p$ is the horizontal axis, and line $q$ is the diagonal $y = x$. Reflect the $M$ first across line $q$, then reflect the result across line $p$. Pick the choice whose picture shows the final image — both the location and the orientation of the letter must match.
Givens: Original $M$ is upright and located in the first quadrant (top-right) at roughly $(0.25, 0.6)$; Line $p$ is horizontal — the $x$-axis; Line $q$ is the diagonal line $y = x$, passing through the origin at $45°$; The reflections are performed in the order: first over $q$, then over $p$; All five answer choices show an $M$ (or rotated $M$) in one of four mini-quadrants together with the same two reference lines
Unknowns: Which mini-figure (A)-(E) shows the correct final position and correct final orientation of the $M$ after the two reflections
Understand
Restated: An upright letter $M$ sits in the first quadrant of a coordinate plane. Line $p$ is the horizontal axis, and line $q$ is the diagonal $y = x$. Reflect the $M$ first across line $q$, then reflect the result across line $p$. Pick the choice whose picture shows the final image — both the location and the orientation of the letter must match.
Givens: Original $M$ is upright and located in the first quadrant (top-right) at roughly $(0.25, 0.6)$; Line $p$ is horizontal — the $x$-axis; Line $q$ is the diagonal line $y = x$, passing through the origin at $45°$; The reflections are performed in the order: first over $q$, then over $p$; All five answer choices show an $M$ (or rotated $M$) in one of four mini-quadrants together with the same two reference lines
Plan
Primary tool: #10 Create a Physical Representation
Secondary: #17 Visualize Spatial Relationships, #3 Eliminate Possibilities, #1 Draw a Diagram
Two reflections are easy to mis-visualize in the head, especially over the diagonal $y = x$. Tool #10 (Physical) wins here: cut a small upright $M$ out of paper, draw lines $p$ and $q$ on a sheet, and physically flip the cutout over $q$, then flip the result over $p$. That removes all guesswork about orientation. Tool #17 (Visualize) is the same move done mentally for the student who's ready. Tool #3 (Eliminate) is the multiple-choice safety net — track the final quadrant first and three of the five options drop out immediately. Tool #1 (Diagram) keeps a clean coordinate sketch of where each intermediate image lands. After the picture work, a one-line check via the "composition of two reflections through intersecting lines $=$ single rotation" theorem confirms the answer.
Execute — Answer: E
5.G.A.1 Step 1 - Sketch the setup.
- Mark line $p$ as the horizontal axis and line $q$ as the diagonal $y = x$.
- Plot the original upright $M$ in the first quadrant — its three peaks point up ($+y$), its two feet sit lower at roughly $(0.05, 0.4)$ and $(0.45, 0.4)$.
- This single picture lets the student track position and orientation through both flips.
💡 Putting $M$ on a coordinate grid turns "reflect over a line" into a concrete picture you can flip — a Grade 5 coordinate-plane move.
8.G.A.3 Step 2 - First reflection: flip over $q$ ($y = x$).
- Reflecting any point over the line $y = x$ swaps its coordinates: $(x, y) \to (y, x)$.
- The center of the $M$ moves from $(0.25, 0.6)$ to $(0.6, 0.25)$ — still in the first quadrant, but now lower-right (closer to the $x$-axis than the $y$-axis).
💡 Swapping $x$ and $y$ is the coordinate fingerprint of a reflection over $y=x$ — Grade 8 transformation rule.
8.G.A.1 Step 3 - Track the orientation through that same flip.
- The $M$'s "up" direction (peaks pointing $+y$) maps to $+x$ — peaks now point to the right.
- Its "left-to-right" baseline (which ran along $+x$) maps to $+y$.
- So the $M$ is now lying on its left side, with peaks pointing right and opening facing left.
- In asymptote notation, that orientation is $\texttt{rotate}(270°) M$ (i.e.
- rotated $90°$ clockwise from upright).
💡 Watching where the peak-arrow goes tells you the new orientation — a Grade 8 reflection-property check.
8.G.A.3 Step 4 - Second reflection: flip the result over $p$ (the $x$-axis).
- Reflection over the $x$-axis negates the $y$-coordinate: $(x, y) \to (x, -y)$.
- So the center at $(0.6, 0.25)$ moves to $(0.6, -0.25)$ — the image leaves the first quadrant and lands in the fourth quadrant (bottom-right).
💡 Reflection over the $x$-axis only flips $y$, leaving $x$ unchanged — straight from the Grade 8 coordinate transformation rules.
8.G.A.1 Step 5 - Track the orientation through this second flip.
- The peak-direction was $(+1, 0)$; reflecting over the $x$-axis keeps $x$ unchanged, so the peaks still point right $(+1, 0)$.
- The opening still faces left.
- So the final $M$ has the same orientation as after Step 1 — $\texttt{rotate}(270°) M$ — but now it sits in the fourth quadrant.
- Match this against the five options: only choice $\textbf{(E)}$ shows $\texttt{rotate}(270°) M$ in the bottom-right (fourth quadrant) of its mini-axes.
💡 Scanning the five mini-pictures and keeping only the one matching both quadrant and orientation is the Grade 8 transformation properties check applied to multiple choice.
5.G.A.1 Sketch the setup. Mark line $p$ as the horizontal axis and line $q$ as the diago 8.G.A.3 First reflection: flip over $q$ ($y = x$). Reflecting any point over the line $y 8.G.A.1 Track the orientation through that same flip. The $M$'s "up" direction (peaks po 8.G.A.3 Second reflection: flip the result over $p$ (the $x$-axis). Reflection over the 8.G.A.1 Track the orientation through this second flip. The peak-direction was $(+1, 0)$ Review
Reasonableness: Quick orientation sanity check on the five choices: (A) $\texttt{rotate}(90°) M$ in Q4 — wrong orientation. (B) $\texttt{rotate}(270°) M$ in Q2 — right orientation, wrong quadrant. (C) $\texttt{rotate}(90°) M$ in Q1 — wrong orientation and wrong quadrant. (D) $\texttt{rotate}(180°) M$ in Q3 — wrong orientation. (E) $\texttt{rotate}(270°) M$ in Q4 — matches both the orientation derived in Steps 3 and 5 and the Q4 position derived in Step 4. The answer is internally consistent: the original $M$ in Q1 ends up in the diagonally-opposite-by-one-quadrant Q4, which is exactly where a single $90°$ clockwise rotation about the origin would send it.
Alternative: Tool #17 (composition shortcut): two reflections over lines through a common point equal a single rotation about that point through twice the directed angle from the first line to the second. Lines $p$ (angle $0°$) and $q$ (angle $45°$) meet at the origin. The directed angle from $q$ to $p$ is $-45°$, so the composition is a rotation by $2 \times (-45°) = -90°$ (i.e. $90°$ clockwise) about the origin. Apply that single rotation to the upright $M$ at $(0.25, 0.6)$: position $(0.25, 0.6) \to (0.6, -0.25)$ in Q4, and orientation upright $\to \texttt{rotate}(270°) M$. Same answer, $\textbf{(E)}$, in one step instead of two.
CCSS standards used (min grade 8)
5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Placing the upright $M$, line $p$ (the $x$-axis), and line $q$ ($y=x$) on a single coordinate grid so the two reflections can be tracked numerically.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Confirming that each reflection preserves the shape of $M$ and only flips its orientation, then verifying via the composition-of-reflections theorem that the two flips equal a $90°$ clockwise rotation about the origin.)8.G.A.3Describe the effect of dilations, translations, rotations, and reflections on coordinates (Applying the coordinate rules $(x, y) \to (y, x)$ for reflection over $y=x$ and $(x, y) \to (x, -y)$ for reflection over the $x$-axis to move the $M$ through both flips.)
⭐ This AMC 8 problem only needs the Grade 8 rule for what reflections do to coordinates — flip the $M$ over $y=x$ (swap $x$ and $y$), then over the $x$-axis (flip the sign of $y$), and you land on (E)!
⭐ This AMC 8 problem only needs the Grade 8 rule for what reflections do to coordinates — flip the $M$ over $y=x$ (swap $x$ and $y$), then over the $x$-axis (flip the sign of $y$), and you land on (E)!
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